Week 8 - Magnetic Field and Magnetic Forces
Ampere was the Newton of Electricity.
James C. Maxwell
Exercise 8.1: Discussion Questions
a) Can a charged particle move trough a magnetic field without experiencing any force? If so, how? If
not, why not?
Solution:
Yes it can. Looking at the force law
Fb = qv × B,
(1)
the magnetic force is zero if v is parallel to B. I.e. the magnetic force is zero if a charged particle
travel along the field lines.
Particles with nonzero components of velocity at nonparallel with B end up spiralling around the
field lines.
b) If the magnetic field do no work, how can it have a significant effect on a particles motion? Are there
any other examples for which a force has a significant effect on the motion without doing any work?
Solution:
The answer is that it can, without doing any work, change the direction of the velocity and therefore
it’s motion. If the particle only has changes to it’s velocity and not it’s speed the kinetic energy does
not change and thus no work is done.
Another example of this is the familiar normal force from mechanics. The normal force on a roller
coaster cart exerted by the rails does no work, but it certainly does affect the roller coaster carts
motion!
c) A student claims that if lightning strikes a metal flag pole, the force exerted by the earths magnetic
field on the current flowing trough the pole can be enough to bend it. Typical lightning currents are
of the order of 104 to 105 amperes. Is the student’s opinion justified?
Solution:
This calls for a rough estimate. The magnetic force magnitude on a straight rod is given by F = ILB,
so assuming the worst possible scenario of a 105 amperes current and a 10 m rod we get a force on
the order of
F ∼ 105 A × 10 m × 10−5 T ∼ 10 N.
1
(2)
A force of the order of 10 N is not negigble. Remember this was just a rough estimate, so this calls
for a more detailed calculation taking the material of the rod into account etc. The student could be
right.
d) A loudspeaker should never be placed in next to a computer monitor or a TV screen. Why not?
Solution:
Speakers contain powerful magnets which can permanently mess up the picture on a CRT dispaly. On
an LCD or plasma display, the effect is less. However, even for the latter two types, there is chance
that something would be magentized.
Exercise 8.2: Gauss’ Law for Magnetism
In a certain region of space, the magnetic field B is not uniform. The magnetic field has both a zcomponent and a component that points radially away from the z-axis. The z-component is given by
Bz (z) = βz, where β > 0. The radial component Br depends only on r, the radial distance from the
z-axis.
a) Use Gauss’ law for magnetism to find the radial component Br as a function of r. Hint: Try a
clindrical Gaussian surface of radius r concentric with the z-axis with one end at z = z0 and the
other at z.
Solution:
We use the hint and align a gaussian cylinder coaxial with the z-axis. This is shown in figure 1.
Now as also shown in this figure the B-field can be decomposed in it’s z-component and it’s radial
component and only the radial component Br will contribute to the flux trough the surface of the
cylinder (not including the top and bottom surface where only the z-component will contribute to
the flux). Further more since the radial field only depend on r it will be uniform along this surface
and this is also the case for the bottom and top surface of the cylinder.
We compute the flux
I
B · dA = 2πr(z − z0 )Br + βzπr2 − βz0 πr2 = 2πr(z − z0 )Br + βπr2 (z − z0 )
(3)
where the last two terms corresponds to the flux from the top and bottom suface. Now by Gauss’
law for magnetism, this has to be zero, so we can solve for Br which yields
1
Br = − βr.
2
(4)
1
Br = − βr.
2
(5)
Answer:
Week 8 – October 24, 2011
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compiled October 22, 2012
Figure 1
Exercise 8.3: Mass Spectrograph
A mass spectrograph is used to measure the masses of ions, or to separate ions of different masses. In
one design for such an instrument, ions with mass m and charge q are accelerated trough a potential
difference V . They then enter a uniform magnetic field that is perpendicular to their velocity, and
they are deflected in a semicicular path of radius R. A detector measures where the ions complete the
semicircle and from this it is easy to cacluate R. The situation where a source S is emitting ions is shown
in figure 2. In the red region there is a potential difference, while in the blue region there is a uniform
magnetic field.
a) Derive the equation for calculating the mass of the ion from measurements of B, V , R and q.
Solution:
The speed of the ions when they enter the region of a uniform magnetic field is found by energy
conservation;
1
qV = mv 2 ⇒ v =
2
Week 8 – October 24, 2011
3
r
2qV
.
m
(6)
compiled October 22, 2012
Figure 2
When the ion enter this region it will feel a magnetic force
Fb = qv × B.
(7)
Now this force is always perpendicular to v, so the acceleration is always perpendicular to v. This
is a characteristic of circular motion, so that the ion will start going in a circle at once it enters the
region of uniform B. Furthermore B is always perpendicular to v so the magnitude is
Fb = qvB = ma = m
v2
.
R
(8)
Thus
r
qBR = mv = m
p
2qV
= 2qV m
m
(9)
or equivalently
m=
qB 2 R2
.
2V
(10)
m=
qB 2 R2
.
2V
(11)
Answer:
Week 8 – October 24, 2011
4
compiled October 22, 2012
b) What potential difference V is needed so that singly ionized
0.150 T magnetic field?
12
C atoms will have R = 50.0 cm in a
Solution:
Rearranging the formula found in (b) we find
V =
qB 2 R2
2m
(12)
The mass of Carbon-12 is 12 u and if it’s singly ionized it obtains a charge e. Thus the necessary
potetial is
2
1.6 × 10−19 × (0.150) × 50.0 × 10−2
V = 22.5904 kV.
V =
2 × 12 × 1.66 × 10−27
(13)
2
1.6 × 10−19 × (0.150) × 50.0 × 10−2
V =
V = 22.5904 kV.
2 × 12 × 1.66 × 10−27
(14)
Answer:
c) Suppose the beam consists of a mixture of 12 C and 14 C ions. If V and B have the same values as in
part (b), calculate the separation of these two isotopes at the detector. Do you think that this beam
separation is sufficient for the two ions to be distinguished?
Solution:
Carbon-14 has a mass of 14 u. Again rearranging the formula in (a) and (b) differences in radii of the
two semicircles will be
s
∆R =
2m14 C V
−
qB 2
s
2m12 C V
= 4 cm.
qB 2
(15)
This beam separation should allow for easy separation of the two ions.
Answer:
∆R = 4 cm
(16)
.
Exercise 8.4: An Electromagnetic Rail Gun
A conducting bar with mass m and length L slides over horizontal rails that are connected to a voltage
source. The voltage source maintains a constant current I in the rails and bar, and a constant, uniform,
Week 8 – October 24, 2011
5
compiled October 22, 2012
Figure 3
vertical magnetic field B fills the region between the rails. This is shown in figure 3 where the black
arrows indicate the direction of the current trough the grey rails and the purple conducting bar, while
the blue arrows (pointing out of the page) indicate the direction of the B-field.
a) Find the magnitude and direction of the net force on the conducting bar. Ignore friction, air resistance,
and electrical resitance.
Solution:
Let’s align the axis of the railgun along the x-axis. Then the force on the conducting bar is then
given by
FB = IL × B = ILB î
(17)
i.e. it is directed along the axis of the rail gun. Since we are ignoring all other forces this is the only
force and thus the net force on the bar.
Answer:
FB = ILB î.
(18)
b) If the bar has mass m, find the distance d that the bar must move along the rails from rest to attain
a speed v.
Solution:
The force found in (a) will provide a constant acceleration given by a = ILB/m. Now for constant
acceleration the familiar relation between distance d and speed v is given by
2ad = v 2 − v02 .
(19)
So starting from rest we get that distance needed to acquire a speed v is
d=
2mv 2
v2
=
.
2a
ILB
(20)
Thus it would be good to go for a huge current and magnetic field when designing a rail gun. The
length L can not be made that large for pracitcal purposes in addition to all the air resistance it
would cause and the greater mass.
Week 8 – October 24, 2011
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Answer:
d=
2mv 2
.
ILB
(21)
c) It has been suggested that rail guns based on this prinicple could accelerate payloads into earth orbit
or beyond. Find hte distance the bar must travel along the rail if it is to reach the escape speed for
the eart (11.2 km/s). Let B = 1.0 T, I = 3.0 kA, m = 1.0 kg and L = 1.0 m. For simplicity assume
the net force on the object is equal to the magnetic force, as in parts (a) and (b), even though gravity
plays an important role in an actual launch into space.
Solution:
This would require a distance d of
2
1.0 × 11.2 × 103
m = 21281.6 m
d=
2 × 3.0 × 103 × 1.0 × 1.0
(22)
In other words a pretty long rail gun. But we have achieved more favourable numbers. The highest
ever magnetic field produced in a lab is about 100 T and and we have certainly produced electrical
currents of about 105 A. With these numbers our distance would be
d = 6.4 m
(23)
which suggests creating such a rail gun MIGHT be possible?
Exercise 8.5: Magnetic Moment of the Hydrogen Atom
In the Bohr model of the hydrogen atom, in the lowest energy state the electron orbits the proton at a
speed of 2.2 × 106 m/s in a circular orbit of radius a0 = 5.3 × 10−11 m called the Bohr radius.
a) What is the orbital period of the electron?
Solution:
Let r = a0 be the radius of the electron and let v be it’s velocity. The period must be the time for
one revolution. Now the distance traveled for a revolution is the circumference 2πr, so the velocity
of the electron can be expressed as
v=
2πr
T
(24)
T =
2πr
.
v
(25)
such that the period is given by
Week 8 – October 24, 2011
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compiled October 22, 2012
Answer:
T =
2πr
.
v
(26)
b) If the orbiting electron is considered to be a current loop, what is the current I?
Solution:
The current is the charge of the electron divided by the period.
I=−
ev
2πr
(27)
I=−
ev
2πr
(28)
Answer:
c) What is the magnetic moment of the atom due to the motion of the electron? Show that it is
proportional to the orbital angular momentum of the electron.
Solution:
The magnetic moment is given by µ = IA where A is the area of the ’current loop’ represented by
the electron orbit. Therfore
ev evr
πr2 = −
µ= −
.
2πr
2
(29)
Now for an electron moving in a circuiar radius the angular momentum is given by L = rme v such
that
µ=−
eL
.
2me
(30)
Now this result turns out to be out by a factor 2 from the actual value which is obtained by relativistic
quantum mechanics. In fact the experimental and theoretical agreement on this value taking quantum
mechanics in to account is a staggering 14 desimal places!
Answer:
µ=−
Week 8 – October 24, 2011
eL
.
2me
8
(31)
compiled October 22, 2012
Exercise 8.6: Force on a Semicircular Wire
Figure 4 shows a conducting wire which enters a region of a uniform magnetic field B pointing out of
the page. In this region the wire takes the form of a semicirle. The direction of the current running
trough it is shown by the arrows. Find the net force on the conducting wire.
Figure 4
Solution:
Figure 5
The wire only feels a force in the region where there is a magnetic field. An infinitesimal segment of
wire feels the force dF = Idl × B. Now as shown in figure 5 the contribution to the y-component of
this force is given by
dFy = |dF| sin θ
(32)
where |dF| = IBdl = IBRdθ so that the magnitude of the entire y-component of force is given by
Z
Fy =
π
Z
dFy = IBR
sin θdθ = 2IBR.
(33)
0
The x-component is given by
Z
π
Fx = IBR
cos θdθ = 0
(34)
0
which we also could have seen from the symmetry the problem. Furhermore by the right hand rule
this force must point in the negative y-direction. Therefore we conclude that
Week 8 – October 24, 2011
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compiled October 22, 2012
F = −2IBRĵ.
(35)
F = −2IBRĵ.
(36)
Answer:
Week 8 – October 24, 2011
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compiled October 22, 2012